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I'm shocked that there isn't a satisfactory answer on this site yet! This can be answered in any number of dimensions with some relatively simple vector math. As follows intuitively (rigorously in the center-of-mass frame), for equal mass particles, the relative velocities of the particles are reversed along the normal direction. All that needs to be done ...


0

The conservation of momentum is simply a statement of Newton's third law of motion. During a collision the forces on the colliding bodies are always equal and opposite at each instant. These forces cannot be anything but equal and opposite at each instant during collision. Hence the impulses (force multiplied by time) on each body are equal and opposite at ...


0

The conservation of momentum is simply a statement of Newton's third law of motion. During a collision the forces on the colliding bodies are always equal and opposite at each instant. These forces cannot be anything but equal and opposite at each instant during collision. Hence the impulses (force multiplied by time) on each body are equal and opposite at ...


1

You didn't cancel out the mass $m$ properly. $$-mv_0\cos a+0=mv_1\cos b+0\\ -v_0\cos a=v_1\cos b\\ \frac{-v_0\cos a}{\cos b}=v_1$$ The incident angle is equal to the exit angle in such collision, $a=b$. The above reduces to: $$v_1=\frac{-v_0\cos a}{\cos b}=\frac{-v_0\cos b}{\cos b}=-v_0$$ And here you see what you probably expected. In the perpendicular ...


0

Your equation applies to a perfect fluid that has no viscosity. N-body simulations are for dark matter only and the dark matter is generally assumed to only interact gravitationally. There are hydrodynamical simulations that include baryons but those are more computationally intensive and so generally done for smaller simulations. If you are just interested ...


2

If the collision is not perfectly along the line connecting the centers of mass of the pucks, they will exert torques on each other as well as forces. The angular momentum of the pair will be conserved, so if the incoming puck was not spinning, the pucks will exit the collision spinning in opposite directions. If the surface they slide on is frictionless, ...


0

Let me summarize the discussion in the comment section. First of all you can safely use Newtons equation of motion $$ F = m g $$ with a constant gravitational acceleration $g$. This is simply because the height of a 10 stories building, lets say $30 m$, is still very very small compared to the radius of the earth which is about $6000 km$. Therefore, as ...


4

Yes, $F=ma$, but also $v=at$. That means that, as you fall for a longer time, your speed will increase. After 1 second, you are going at $9.8 m/s$ or $35 km/h$, about the speed of Usain Bolt. After 10 seconds you would reach $98 m/sec$ or $350 km/h$. For a free-falling human, the air resistance actually limits you to about $200 km/h$. When you hit the ...


-2

F=ma is saying that if a mass 'm kg' is accelerated at 'a metres per second per second' then there has to be a constant force of 'F' pushing it. a better equation to represent the effects of gravity would be: F=(G*m1*m2)/d^2 where: G= universal gravitational constant (6.6738410^-11 m^3 kg^(-1) s^(-2)), g= acceleration due to gravity (9.81 m s^(-1) ...


2

Remember that the rest mass of a proton is about 940MeV. So, a proton at 0.6c is pretty darn energetic. I will leave a precise energy up to the reader. However, there are a number of experiments in the physics literature of smashing protons into various things. At low (1MeV-ish) energies, you are looking mainly at classic Rutherford scattering cross ...


1

As the Earth isn't a closed system with regard to other objects in the Solar System, including atomic particles, its momentum is affected by collisions. But if the collisions were inelastic (if the atomic particles were absorbed into elements of the crust and the atmosphere), the momentum of the atomic particles and the momentum of the Earth involved in the ...


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It does affect the Earth, but at a rate so slow that the sun will expand into a red giant before that happens.


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Game engines typically do approximations that may or may not relate in anyway to reality. For example CORs typically reduce with increasing velocity, but that's typically left out of games. I've seen game engines just take an unweighted average CORs of two impacting objects. A physically more realistic approach would be to do a weight the average of the CORs ...


0

Kinetic energy is the work required to accelerate a mass from rest to a velocity (KE = 1/2 mv^2). Momentum is a measure of the amount of movement a mass has at a velocity (p = mv). Kinetic energy may be considered a process, and momentum may be considered the result of a process. Momentum is conserved, but kinetic energy seems to come and go, as it ...


0

That's easy. Think in a simple example that this happens. Imagine two particles of equal masses moving at $\vec v_1 = \vec v$ and $\vec v_2 = -\vec v$. Their momentum: $\vec p_1 = m\vec v_1$ and $\vec p_2 = m\vec v_2$. The momentum of the system is therefore: $$ \vec p = \vec p_1 + \vec p_2 = m\vec v_1 + m\vec v_2 = m\vec v - m\vec v = \vec 0 $$ The ...


0

The answer by RGJ is not exact: the ball will always travel back up eh. The only difference is that:if the collision is inelastic $e<1$, if the collision is elastic $e=1$, But as correctly said by angel, it is almost impossible that this may happen and the coefficient of restitution will be at most 0.99...... This because e is the ...


3

Find the nearest box-spring mattress. With your hand, execute a slow motion "impact" between your hand and the mattress. You should notice that when your hand is not touching the mattress, there is, of course, no force between your hand and the mattress. As your hand begins to touch the mattress, you should feel a very light force. As your hand presses ...


14

Elastic collisions do happen at the LHC. The TOTEM experiment measures the differential cross section (rate as a function of angle) for proton-proton elastic scattering at the LHC. Here is their latest result. They don't publish an estimate of the elastic cross section, but according to their data it must be at least 25 mb (millibarns) (my first version of ...


8

Anything that is not forbidden must happen. That's an important statement to keep in mind when approaching quantum physics. It doesn't mean that anything that can happen always happens, but it must happen at some time or another just like someone eventually has to win the lottery. That said, some protons do go through the LHC, ram into each other and ...


6

elementary particles (e.g. protons) Protons aren't elementary particles, they're made of partons (quarks and gluons) in "soup". Below, $\lambda$ is the wavelength corresponding to the energy of the interaction via the usual de Broglie relation and $r_p$ is the radius of the proton. At low energy with $\lambda >> r_p$ the interactions are just ...


1

If we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether and use a single conservation-of-energy equation... The radius of a ball is always irrelevant to the outcome of a collision, what counts is the ratio of the masses. Two equations are always necessary to determine the outcome ...


4

In both cases and at all times, the force from the (wall/tire) on the hammer equals the force from the hammer on the (wall/tire) : total momentum must be conserved. However, in the first case, the initial energy is dissipated in the wall (as heat and/or damage), so at the end the hammer is stopped. In the second case the initial energy is stored as ...


2

Here's how. The ball gains a velocity $v$ due to gravity before hitting the ground. So each time it hits the ground its velocity is changed from $v$ to $-v$ (taking down as positive) during the collision, then returning again with $v$. The force $F_1= m\frac {dv}{dt}$ is experienced by the ball due to the collision, however this force is felt after the ...


0

Yes. Some of the impulse force is used in work which rotates the rod, and some of the impulse force is used in work which changes the bullet's trajectory. The distribution of impulse force between the rod and the bullet is equal.


5

The problem does not mention any radii, but if we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether The accepted answer has confused you: You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore ...


2

The ship and cannonballs will both have the same velocity in the end. The easiest way to think about this from a reference frame in which the ship is stationary before it intersects the cannonballs. The cannonballs will be moving toward the ship in this frame. When the cannonballs reach the springs on the ship their kinetic energy will be converted into ...



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